Question

The hybridization of Ni in $\text{Ni}{\left(\text{CO}\right)}_{\text{4}}$ is

• A. $\text{s}{\text{p}}^{}$
• B. $\text{s}{\text{p}}^{\text{2}}$
• C. $\text{s}{\text{p}}^{\text{3}}d$
• D. $\text{s}{\text{p}}^{\text{3}}$

Answer 1

The correct option is D $\text{s}{\text{p}}^{\text{3}}$

In $\left[Ni\right(CO{)}_4]$, the valence shell electronic configuration of ground state Ni atom is $3d^{8}4s^2$. All of these 10 electrons are pushed into 3d orbitals and get paired up when strong field CO ligands approach $Ni$ atom. The empty 4s and three 4p orbitals undergo $s{p}^3$ hybridization and form bonds with four CO ligands to give $\left[Ni\right(CO{)}_4]$ with tetrahedral geometry. Since all the electrons are paired, $\left[Ni\right(CO{)}_4]$ is diamagnetic.

Hence, option D is correct.